Modular Exponentiation (Power in Modular Arithmetic)
Last Updated :
14 Jul, 2025
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Given three integers x, n, and M, compute (xn) % M (remainder when x raised to the power n is divided by M).
Examples :
Input: x = 3, n = 2, M = 4
Output: 1
Explanation: 32 % 4 = 9 % 4 = 1.Input: x = 2, n = 6, M = 10
Output: 4
Explanation: 26 % 10 = 64 % 10 = 4.
Table of Content
[Naive Approach] Repeated Multiplication Method - O(n) Time and O(1) Space
We initialize the result as 1 and iterate from 1 to n, updating the result by multiplying it with x and taking the modulo by M in each step to keep the number within integer bounds.
#include<iostream>
using namespace std;
int powMod(int x, int n, int M) {
// Initialize result as 1 (since anything power 0 is 1)
int res = 1;
// n times to multiply x with itself
for(int i = 1; i <= n; i++) {
// Multiplying res with x
// and taking modulo to avoid overflow
res = (res * x) % M;
}
return res;
}
int main() {
int x = 3, n = 2, M = 4;
cout << powMod(x, n, M) << endl;
return 0;
}
public class GfG {
public static int powMod(int x, int n, int M) {
int res = 1;
for (int i = 1; i <= n; i++) {
// Multiplying res with x and
// taking modulo to avoid overflow
res = (res * x) % M;
}
return res;
}
public static void main(String[] args) {
int x = 3, n = 2, M = 4;
System.out.println(powMod(x, n, M));
}
}
def powMod(x, n, M):
res = 1
# loop from 1 to n
for _ in range(n):
# Multiplying res with x
# and taking modulo to avoid overflow
res = (res * x) % M
return res
if __name__ == "__main__":
x, n, M = 3, 2, 4
print(powMod(x, n, M))
using System;
class GfG {
static int powMod(int x, int n, int M) {
int res = 1;
for (int i = 1; i <= n; i++) {
// Multiplying res with x and
// taking modulo to avoid overflow
res = (res * x) % M;
}
return res;
}
static void Main() {
int x = 3, n = 2, M = 4;
Console.WriteLine(powMod(x, n, M));
}
}
function powMod(x, n, M) {
let res = 1;
// Loop n times, multiplying x and taking modulo at each step
for (let i = 1; i <= n; i++) {
// Multiplying res with x and
// taking modulo to avoid overflow
res = (res * x) % M;
}
return res;
}
// Driver Code
let x = 3, n = 2, M = 4;
console.log(powMod(x, n, M));
Output
1
[Expected Approach] Modular Exponentiation Method - O(log(n)) Time and O(1) Space
The idea of binary exponentiation is to reduce the exponent by half at each step, using squaring, which lowers the time complexity from O(n) to O(log n).
-> xn = (xn/2)2 if n is even.
-> xn = x*xn-1 if n is odd.
Step by step approach:
- Start with the result as 1.
- Use a loop that runs while the exponent
nis greater than 0. - If the current exponent is odd, multiply the result by the current base and apply the modulo.
- Square the base and take the modulo to keep the value within bounds.
- Divide the exponent by 2 (ignore the remainder).
- Repeat the process until the exponent becomes 0.
#include<iostream>
using namespace std;
int powMod(int x, int n, int M) {
int res = 1;
// Loop until exponent becomes 0
while(n >= 1) {
// n is odd, multiply result by current x and take modulo
if(n & 1) {
res = (res * x) % M;
// Reduce exponent by 1 to make it even
n--;
}
// n is even, square the base and halve the exponent
else {
x = (x * x) % M;
n /= 2;
}
}
return res;
}
int main() {
int x = 3, n = 2, M = 4;
cout << powMod(x, n, M) << endl;
}
class GfG {
public int powMod(int x, int n, int M) {
int res = 1;
// Loop until exponent becomes 0
while (n >= 1) {
// n is odd, multiply result by current x and take modulo
if ((n & 1) == 1) {
res = (res * x) % M;
// Decrease n to make it even
n--;
} else {
// n is even, square the base and halve the exponent
x = (x * x) % M;
n /= 2;
}
}
return res;
}
public static void main(String[] args) {
int x = 3, n = 2, M = 4;
GfG obj = new GfG();
System.out.println(obj.powMod(x, n, M));
}
}
def powMod(x, n, M):
res = 1
# Loop until exponent becomes 0
while n >= 1:
# n is odd, multiply result by current x and take modulo
if n % 2 == 1:
res = (res * x) % M
# Make n even
n -= 1
else:
# n is even, square the base and halve the exponent
x = (x * x) % M
n //= 2
return res
if __name__ == "__main__":
x, n, M = 3, 2, 4
print(powMod(x, n, M))
using System;
class GfG {
public int powMod(int x, int n, int M) {
int res = 1;
// Loop until exponent becomes 0
while (n >= 1) {
// n is odd, multiply result by current x and take modulo
if ((n & 1) == 1) {
res = (int)((1L * res * x) % M);
// Reduce exponent by 1
n--;
} else {
// n is even, square the base and halve the exponent
x = (int)((1L * x * x) % M);
n /= 2;
}
}
return res;
}
public static void Main() {
int x = 3, n = 2, M = 4;
GfG obj = new GfG();
Console.WriteLine(obj.powMod(x, n, M));
}
}
function powMod(x, n, M) {
let res = 1;
// Loop until exponent becomes 0
while (n >= 1) {
// If n is odd, multiply result by current x and take modulo
if (n % 2 === 1) {
res = (res * x) % M;
n -= 1;
} else {
// If n is even, square the base and halve the exponent
x = (x * x) % M;
n /= 2;
}
}
return res;
}
// Driver Code
let x = 3, n = 2, M = 4;
console.log(powMod(x, n, M));
Output
1
